Question

ii) PQRS is a rectangle. P(a, a-1), Q( a - 3, a +1), R (-2a - 1,- a -1) and
S(-. - 2a - 1) are the vertices of D PQRS. Find the value of 'a' and hence find the
co-ordinates of vertices of PQRS.​

Answer 1

Answer:hi

Step-by-step explanation:

Since it is a parallelogram (rectangle), we have the following property:

x_1+x_3=x_2+x_4\\ Hence \: a+ (-2a-1)-(-a)=a-3 \\ Solving, \\ -1+3=a\\ a=2

Answer 2

Answer:

opposite sides of rectangle are congruent hence distance PQ is equal to distance QR and distance QR is equal to distance RS so it is easy to find the value of a

first find the distance PQ and distance QR

opposite sides of rectangular contract hence show that distance PQ is equal to distance QR

hope you will understand